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17 votes
17 votes
A 3520 kg truck moving north at

18.5 m/s makes an INELASTIC
collision with an 1480 kg car
moving east. After colliding, they
have a velocity of 13.6 m/s at
72.6°. What was the initial velocity
of the car?

User Kirk
by
3.3k points

2 Answers

27 votes
27 votes

Answer:

13.7 m/s

Explanation:

User Mlucy
by
3.0k points
14 votes
14 votes

Answer: 13.7 m/s

Step-by-step explanation:


$$Mass of $T r u c k$$ \ \left(m_(T)\right)=3520\mathrm{kg}


V_{T_(y)}=18.5 \mathrm{~m} / \mathrm{s} \\&V_{\text {Tix }}=0 \mathrm{~m} / \mathrm{s}


\mathrm{Mass \ of \ car \ (m_c) = 1480 \ kg}


V_(c i x)=\text { ? ; } V_(c i y)=0 \mathrm{~m} / \mathrm{s}


Final \ velocity $\left(V_(f)\right)=13.6 \mathrm{~m} / \mathrm{s} \ \ \theta=72.6$


v_(f x)=v_(f) \cos \theta=(13.6) \cos 72.6=4.067 \mathrm{~m} / \mathrm{s}


\mathrm {Using \ the \ conservation \ of \ momentum \ along\ the \ $x$-axis}


\begin{aligned}m_(T) v_(T i x)+m_(c) v_{c_(i x)} &=\left(m_(T)+m_(c)\right) v_(f x) \\0+(1480) V_{c_(i x)} &=(3520+1480)(4.067) \\(1480) v_(c i x) &=20335 \\V_(c i x) &=13.7 \mathrm{~m} / \mathrm{sec}\end{aligned}

User Xiak
by
2.9k points